Well if its between 10 & 20 take the median which is 15 and multiple it by the about of days which is 56 and you get 840 dollars <span />
Answer:
we need the question
Step-by-step explanation:
Answer:
x = 3.4
Step-by-step explanation:
Given the graphs of 2 functions, then the solutions to both are at their points of intersection, that is
where f(x) = g(x)
There are 3 possible points of intersection to f(x) = g(x)
The one on the list , however, is at x = 3.4
To do this problem you would first need to factor out a variable, which in this case I would want to do the first equation because it is isolated. Now the equations would look like this:
x = -2y - 1
4x - 4y = 20
Since we know that x is now equal to -2y - 1 we can plug it in to the x value in the second equation:
4 (-2y -1) - 4y = 20
-8y -4 - 4y
-12y - 4 = 20
-12y = 24
y = -2
Now that we know the y value plug the y value to one equation to find the x, I will be using the first equation
x + 2(-2) = -1
x - 4 = -1
x = 3
Solutions:
y = -2
x = 3
Given : In Right triangle ABC, AC=6 cm, BC=8 cm.Point M and N belong to AB so that AM:MN:NB=1:2.5:1.5.
To find : Area (ΔMNC)
Solution: In Δ ABC, right angled at C,
AC= 6 cm, BC= 8 cm
Using pythagoras theorem
AB² =AC²+ BC²
=6²+8²
= 36 + 64
→AB² =100
→AB² =10²
→AB =10
Also, AM:MN:NB=1:2.5:1.5
Then AM, MN, NB are k, 2.5 k, 1.5 k.
→2.5 k + k+1.5 k= 10
→ 5 k =10
Dividing both sides by 2, we get
→ k =2
MN=2.5×2=5 cm, NB=1.5×2=3 cm, AM=2 cm
As Δ ACB and ΔMNC are similar by SAS.
So when triangles are similar , their sides are proportional and ratio of their areas is equal to square of their corresponding sides.
![\frac{Ar(ACB)}{Ar(MNC)}=[\frac{10}{5}]^{2}](https://tex.z-dn.net/?f=%5Cfrac%7BAr%28ACB%29%7D%7BAr%28MNC%29%7D%3D%5B%5Cfrac%7B10%7D%7B5%7D%5D%5E%7B2%7D)
But Area (ΔACB)=1/2×6×8= 24 cm²[ACB is a right angled triangle]
→ Area(ΔMNC)=24÷4
→Area(ΔMNC)=6 cm²
